Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{-5q + 20}{5q^2 - 15q - 20} \times \dfrac{q^2 + 8q + 7}{q + 9} $
Explanation: First factor out any common factors. $a = \dfrac{-5(q - 4)}{5(q^2 - 3q - 4)} \times \dfrac{q^2 + 8q + 7}{q + 9} $ Then factor the quadratic expressions. $a = \dfrac {-5(q - 4)} {5(q + 1)(q - 4)} \times \dfrac {(q + 1)(q + 7)} {q + 9} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-5(q - 4) \times (q + 1)(q + 7) } { 5(q + 1)(q - 4) \times (q + 9)} $ $a = \dfrac {-5(q + 1)(q + 7)(q - 4)} {5(q + 1)(q - 4)(q + 9)} $ Notice that $(q + 1)$ and $(q - 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-5\cancel{(q + 1)}(q + 7)(q - 4)} {5\cancel{(q + 1)}(q - 4)(q + 9)} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $a = \dfrac {-5\cancel{(q + 1)}(q + 7)\cancel{(q - 4)}} {5\cancel{(q + 1)}\cancel{(q - 4)}(q + 9)} $ We are dividing by $q - 4$ , so $q - 4 \neq 0$ Therefore, $q \neq 4$ $a = \dfrac {-5(q + 7)} {5(q + 9)} $ $ a = \dfrac{-(q + 7)}{q + 9}; q \neq -1; q \neq 4 $